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Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6
Binomial Theorem | CBSE | Class 11 | Exercise 8.2 | Maths
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6

The overall term

    \[Tr+1\]

in the binomial extension is given by

    \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\]

Here

    \[a\text{ }=\text{ }1\]

,

    \[b\text{ }=\text{ }x\]

and

    \[n\text{ }=\text{ }m\]

Putting the worth

    \[Tr+1\text{ }=\text{ }m\text{ }Cr\text{ }1m-r\text{ }xr\]

    \[=\text{ }m\text{ }Cr\text{ }xr\]

We need coefficient of

    \[x2\]

∴ putting

    \[r\text{ }=2\]

    \[T2+1\text{ }=\text{ }mC2\text{ }x2\]

The coefficient of

    \[x2\text{ }=\text{ }mC2\]

Considering that coefficient of

    \[x2\text{ }=\text{ }mC2\text{ }=\text{ }6\]

NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 30

    \[\Rightarrow \left( m\text{ }\text{ }1 \right)\text{ }=\text{ }12\]

    \[\Rightarrow m2m\text{ }\text{ }12\text{ }=0\]

    \[\Rightarrow m24m\text{ }+\text{ }3m\text{ }\text{ }12\text{ }=0\]

    \[\Rightarrow \left( m\text{ }\text{ }4 \right)\text{ }+\text{ }3\text{ }\left( m\text{ }\text{ }4 \right)\text{ }=\text{ }0\]

    \[\Rightarrow \left( m+3 \right)\text{ }\left( m\text{ }\text{ }4 \right)\text{ }=\text{ }0\]

    \[\Rightarrow m\text{ }=\text{ }\text{ }3,\text{ }4\]

We need positive worth of m so

    \[m\text{ }=\text{ }4\]

 

i

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